There are only three known solutions to Brocard's problem, and the near misses all seem to occur early on. Notice how, for example, 3! is just 3 shy of a square (compared to 1 shy of a square which is what Brocard's problem asks for). Still, the differences between a factorial and the next higher perfect square don't make for a consistently ascending order sequence. For a few values of $n$ , (such as 4, 7, 10, 24, 26, 42, 117, 135) this difference is smaller than the previous difference. In general, however, the difference between a factorial and the next perfect square widens as $n$ gets larger.

The following table gives the square root of $n!$ to six decimal places, and then the difference between the factorial and the next higher square (obtained by taking the ceiling of the square root of $n!$ and squaring that integer).